Question

San Diego is one of the largest cities in the United States. In 2012, the population of San Diego was 1,309,000. By 2016, the population had grown to 1,375,000, and reached 1,386,000 in 2020.

(a) What was the growth rate (k) from 2012 to 2016?

(b) What was the growth rate (k) from 2016 to 2020?

(c) Compare the growth rate from each period.

(d) If the current growth rate continues, when will the population of San Diego reach 1.5 million?

Answer 1

1,4 million is the correct answer

Explanation:

Answer 2

(a) k = 0.04919 = 4.919% (4 s.f.)

(b) k = 0.007968 = 0.7958% (4 s.f.)

(c) The growth rate from 2016 to 2020 is significantly less.

(d) 2059, so during the 4 year period 2056 - 2060.

Explanation:

Given population in San Diego:

• 2012 = 1,309,000
• 2016 = 1,375,000
• 2020 = 1,386,000

Population Growth Formula

`P=P_0 e^(kt)`

where:

• P = population at time t (in millions)
• P₀ = initial population
• e = Euler's number
• k = relative growth rate
• t = time elapsed

Part (a)

Let the population in 2012 be the "initial population":

`\implies P=1.309e^(kt)`

To find the growth rate (k) from 2012 to 2016, substitute t = 1 (since we want to find the growth rate over a period of 4 years) and P = 1.375 into the equation and solve for k:

`\implies 1.375=1.309e^(k)`

`\implies (1.375)/(1.309)=e^(k)`

`\implies (125)/(119)=e^(k)`

`\implies \ln \left((125)/(119)\right)=\ln e^(k)`

`\implies \ln \left((125)/(119)\right)=k \ln e`

`\implies k=\ln \left((125)/(119)\right)`

`\implies k=0.0491902441...`

`\implies k=0.04919\; \sf (4\;s.f.)`

Therefore, the growth rate for the period of 4 years from 2012 to 2016 was 0.04919 = 4.919% (4 s.f.).

Part (b)

Let the population in 2016 be the "initial population":

`\implies P=1.375e^(kt)`

To find the growth rate (k) from 2016 to 2020, substitute t = 1 (since we want to find the growth rate over a period of 4 years) and P = 1.386 into the equation and solve for k:

`\implies 1.386=1.375e^(k)`

`\implies (1.386)/(1.375)=e^(k)`

`\implies (126)/(125)=e^(k)`

`\implies \ln \left((126)/(125)\right)=\ln e^(k)`

`\implies \ln \left((126)/(125)\right)=k \ln e`

`\implies k=\ln \left((126)/(125)\right)`

`\implies k=0.00796816964..`

`\implies k=0.007968\; \sf (4\;s.f.)`

Therefore, the growth rate for the period of 4 years from 2016 to 2020 was 0.007968 = 0.7968% (4 s.f.).

Part (c)

The growth rate from 2012 to 2016 was 0.04919 = 4.919%

The growth rate from 2016 to 2020 was 0.007968 = 0.7968%

Therefore, as the growth rate from 2016 to 2020 is significantly less than the growth rate from 2012 to 2016, the growth of the population is slowing.

Part (d)

To calculate when the population of San Diego will reach 1.5 million if the current growth rate of 0.007968 continues, substitute P₀ = 1.386, k = 0.007968... and P = 1.5 into the equation and solve for t:

`\implies 1.5=1.386e^{t \ln \left((126)/(125)\right)}`

`\implies (1.5)/(1.386)=e^{t \ln \left((126)/(125)\right)}`

`\implies (250)/(231)=e^{t \ln \left((126)/(125)\right)}`

`\implies \ln \left((250)/(231)\right)=\ln e^{t \ln \left((126)/(125)\right)}`

`\implies \ln \left((250)/(231)\right)=t \ln \left((126)/(125)\right)\ln e`

`\implies \ln \left((250)/(231)\right)=t \ln \left((126)/(125)\right)`

`\implies t=(\ln \left((250)/(231)\right))/(\ln \left((126)/(125)\right))`

`\implies t=9.91987003...`

As each time period is 4 years long:

⇒ 2020 + 9.91987003... × 4 = 2059.67948

So the population of San Diego will reach 1.5 million during the year 2059, and so during the 4 year period 2056 - 2060.